[proofplan] Subtract the two weak equations and apply the energy method to the difference $w := u-v$. The difference has zero initial displacement and zero initial velocity, so the conserved wave energy starts at zero. For energy-class weak solutions, the energy identity is justified by testing the weak equation against time-regularized approximations of $\partial_t w$ and passing to the limit. Since the energy is a sum of nonnegative terms, zero energy forces both $\partial_t w$ and $\nabla w$ to vanish at every time, and the zero Dirichlet condition then gives $w=0$ in $H^1_0(U)$. [/proofplan] [step:Subtract the two weak equations and identify the zero initial data] Define the difference as the map \begin{align*} w: [0,T] \to H^1_0(U) \end{align*} given by $w(t)=u(t)-v(t)$ for each $t \in [0,T]$. Here $\mathcal{L}^n$ denotes $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and $\mathcal{L}^1$ denotes one-dimensional Lebesgue measure on $\mathbb{R}$. Since $u,v \in C([0,T];H^1_0(U)) \cap C^1([0,T];L^2(U))$, we have \begin{align*} w \in C([0,T];H^1_0(U)) \cap C^1([0,T];L^2(U)). \end{align*} By linearity of the weak solution formulation for the homogeneous [wave equation](/page/Wave%20Equation), $w$ is a weak solution of \begin{align*} \partial_t^2 w - \Delta w = 0 \quad \text{in } U \times (0,T) \end{align*} with the homogeneous Dirichlet boundary condition on $\partial U \times (0,T)$. The equality of the initial data gives \begin{align*} w(0)=0 \quad \text{in } H^1_0(U) \end{align*} and \begin{align*} \partial_t w(0)=0 \quad \text{in } L^2(U). \end{align*} [/step] [step:Derive the energy identity for the difference] Define the wave energy map \begin{align*} E_w: [0,T] \to [0,\infty) \end{align*} by \begin{align*} E_w(t)=\frac{1}{2}\int_U |\partial_t w(x,t)|^2 \, d\mathcal{L}^n(x) + \frac{1}{2}\int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x) \end{align*} for each $t \in [0,T]$. The two integrals are finite because $\partial_t w(t) \in L^2(U)$ and $w(t) \in H^1_0(U)$. For smooth solutions, testing the equation against $\partial_t w$ and integrating over $U$ gives \begin{align*} \frac{d}{dt}E_w(t)=0. \end{align*} For the present energy-class weak solution, choose a standard one-dimensional mollifier in time and let $w_\varepsilon$ denote the corresponding time-regularization of $w$ on compact subintervals of $(0,T)$. Endpoint issues are avoided by first working on an arbitrary closed interval $[a,b] \subset (0,T)$ and choosing $\varepsilon < \operatorname{dist}([a,b],\{0,T\})$; only after the interior identity is obtained do we pass to endpoints using the continuity of $E_w$. Precisely, the weak formulation means that for every [test function](/page/Test%20Function) $\phi \in C_c^\infty((0,T);H^1_0(U))$, \begin{align*} \int_0^{\mathsf T} (\partial_t w(t),\partial_t \phi(t))_{L^2(U)} \, d\mathcal{L}^1(t) - \int_0^{\mathsf T} \int_U \nabla w(x,t)\cdot \nabla \phi(x,t) \, d\mathcal{L}^n(x)\, d\mathcal{L}^1(t)=0. \end{align*} Let $H^{-1}(U)$ denote the [dual space](/page/Dual%20Space) $(H^1_0(U))^*$. After time-regularization on an interior interval, $\partial_t w_\varepsilon(t) \in H^1_0(U)$ for each interior time $t$, and the regularized equation holds as an identity in $H^{-1}(U)$. Evaluating that $H^{-1}(U)$-valued identity on $\partial_t w_\varepsilon(t)$ gives the pointwise-in-time energy derivative. Integrating in time gives $E_{w_\varepsilon}(t)=E_{w_\varepsilon}(s)$ for all $s,t \in [a,b]$. Passing to the limit as $\varepsilon \to 0$ in the spaces $C([a,b];H^1_0(U))$ and $C([a,b];L^2(U))$ yields \begin{align*} E_w(t)=E_w(s) \end{align*} for all $s,t \in [a,b]$. Since $[a,b] \subset (0,T)$ was arbitrary and $E_w$ is continuous on $[0,T]$, the same identity holds for all $s,t \in [0,T]$. [guided] The formal energy computation uses $\partial_t w$ as a test function, but a weak solution need not have enough time regularity for this test to be immediately legitimate. The standard way to justify the computation is to smooth only in the time variable. Let $\rho \in C_c^\infty((-1,1))$ be a nonnegative function satisfying \begin{align*} \int_{\mathbb{R}} \rho(r) \, d\mathcal{L}^1(r)=1, \end{align*} and for $\varepsilon > 0$ define the map \begin{align*} \rho_\varepsilon: \mathbb{R} \to \mathbb{R} \end{align*} by $\rho_\varepsilon(r)=\varepsilon^{-1}\rho(r/\varepsilon)$ for each $r \in \mathbb{R}$. On a compact interval $[a,b] \subset (0,T)$ with $\varepsilon < \operatorname{dist}([a,b],\{0,T\})$, define the time-regularized function \begin{align*} w_\varepsilon: [a,b] \to H^1_0(U) \end{align*} by \begin{align*} w_\varepsilon(t)=\int_{\mathbb{R}} \rho_\varepsilon(t-s)w(s) \, d\mathcal{L}^1(s) \end{align*} for each $t \in [a,b]$, where $w$ is extended slightly outside $[0,T]$ by any fixed energy-class extension. Since convolution in time commutes with the time derivative on the interior interval, $w_\varepsilon$ is smooth as a map into $H^1_0(U)$ and $\partial_t w_\varepsilon$ is an admissible time-regularized test function. The weak equation for $w$ says that for every $\phi \in C_c^\infty((0,T);H^1_0(U))$, \begin{align*} \int_0^{\mathsf T} (\partial_t w(t),\partial_t \phi(t))_{L^2(U)} \, d\mathcal{L}^1(t) - \int_0^{\mathsf T} \int_U \nabla w(x,t)\cdot \nabla \phi(x,t) \, d\mathcal{L}^n(x)\, d\mathcal{L}^1(t)=0. \end{align*} Let $H^{-1}(U)$ denote the dual space $(H^1_0(U))^*$. Equivalently, $\partial_t^2 w - \Delta w = 0$ holds as an $H^{-1}(U)$-valued distribution in time. Convolving this distributional identity in the time variable with $\rho_\varepsilon$ is legitimate on $[a,b]$ because the support of $\rho_\varepsilon(t-\cdot)$ stays inside $(0,T)$ for $t \in [a,b]$. Since time convolution commutes with $\partial_t$ and with the spatial weak gradient, the regularized function satisfies \begin{align*} \partial_t^2 w_\varepsilon - \Delta w_\varepsilon = 0 \end{align*} as an identity in $H^{-1}(U)$ for each $t \in [a,b]$. Because $w \in C^1([0,T];L^2(U))$, the time-convolution satisfies $\partial_t w_\varepsilon \in C^1([a,b];L^2(U))$ and \begin{align*} \partial_t^2 w_\varepsilon(t)=\int_{\mathbb{R}} \partial_t\rho_\varepsilon(t-s)\partial_t w(s) \, d\mathcal{L}^1(s) \end{align*} as an $L^2(U)$-valued Bochner integral for each $t \in [a,b]$. Thus the $H^{-1}(U)$ pairing of $\partial_t^2 w_\varepsilon(t)$ with $\partial_t w_\varepsilon(t)$ is the $L^2(U)$ [inner product](/page/Inner%20Product). Evaluating the regularized identity on the test vector $\partial_t w_\varepsilon(t) \in H^1_0(U)$ gives \begin{align*} \int_U \partial_t^2 w_\varepsilon(x,t)\partial_t w_\varepsilon(x,t) \, d\mathcal{L}^n(x) + \int_U \nabla w_\varepsilon(x,t)\cdot \nabla \partial_t w_\varepsilon(x,t) \, d\mathcal{L}^n(x)=0 \end{align*} for $t \in [a,b]$. The first integral is the time derivative of the kinetic energy: \begin{align*} \int_U \partial_t^2 w_\varepsilon(x,t)\partial_t w_\varepsilon(x,t) \, d\mathcal{L}^n(x)=\frac{d}{dt}\left(\frac{1}{2}\int_U |\partial_t w_\varepsilon(x,t)|^2 \, d\mathcal{L}^n(x)\right). \end{align*} The second integral is the time derivative of the potential energy: \begin{align*} \int_U \nabla w_\varepsilon(x,t)\cdot \nabla \partial_t w_\varepsilon(x,t) \, d\mathcal{L}^n(x)=\frac{d}{dt}\left(\frac{1}{2}\int_U |\nabla w_\varepsilon(x,t)|^2 \, d\mathcal{L}^n(x)\right). \end{align*} Therefore the regularized energy map \begin{align*} E_{w_\varepsilon}: [a,b] \to [0,\infty) \end{align*} defined by \begin{align*} E_{w_\varepsilon}(t)=\frac{1}{2}\int_U |\partial_t w_\varepsilon(x,t)|^2 \, d\mathcal{L}^n(x) + \frac{1}{2}\int_U |\nabla w_\varepsilon(x,t)|^2 \, d\mathcal{L}^n(x) \end{align*} for each $t \in [a,b]$ is constant on $[a,b]$. Finally, time mollification converges strongly in the energy topology: \begin{align*} w_\varepsilon \to w \quad \text{in } C([a,b];H^1_0(U)) \end{align*} and \begin{align*} \partial_t w_\varepsilon \to \partial_t w \quad \text{in } C([a,b];L^2(U)). \end{align*} Passing to the limit in the two energy terms gives \begin{align*} E_w(t)=E_w(s) \end{align*} for all $s,t \in [a,b]$. Since $[a,b]$ was an arbitrary compact subinterval of $(0,T)$ and $E_w$ is continuous on $[0,T]$, the identity extends to all $s,t \in [0,T]$. [/guided] [/step] [step:Use the zero initial energy to force both energy terms to vanish] Evaluating the energy identity at $s=0$ gives \begin{align*} E_w(t)=E_w(0) \end{align*} for every $t \in [0,T]$. Since $w(0)=0$ in $H^1_0(U)$ and $\partial_t w(0)=0$ in $L^2(U)$, \begin{align*} E_w(0)=0. \end{align*} Thus \begin{align*} E_w(t)=0 \end{align*} for every $t \in [0,T]$. Both terms in $E_w(t)$ are nonnegative. Hence, for every $t \in [0,T]$, \begin{align*} \int_U |\partial_t w(x,t)|^2 \, d\mathcal{L}^n(x)=0 \end{align*} and \begin{align*} \int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x)=0. \end{align*} Therefore \begin{align*} \partial_t w(t)=0 \quad \text{in } L^2(U) \end{align*} and \begin{align*} \nabla w(t)=0 \quad \text{in } L^2(U;\mathbb{R}^n) \end{align*} for every $t \in [0,T]$. [/step] [step:Use the Dirichlet condition to conclude that the difference vanishes] Fix $t \in [0,T]$. Since $w(t) \in H^1_0(U)$ and $\nabla w(t)=0$ in $L^2(U;\mathbb{R}^n)$, the [Poincare inequality](/theorems/63) for $H^1_0(U)$ on the bounded smooth domain $U$ gives \begin{align*} \int_U |w(x,t)|^2 \, d\mathcal{L}^n(x) \leq C_U^2 \int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x)=0, \end{align*} where $C_U>0$ is the Poincare constant of $U$. Hence $w(t)=0$ in $L^2(U)$. Together with $\nabla w(t)=0$ in $L^2(U;\mathbb{R}^n)$, this gives \begin{align*} w(t)=0 \quad \text{in } H^1_0(U) \end{align*} for every $t \in [0,T]$. Since $w(t)=u(t)-v(t)$ and $\partial_t w(t)=\partial_t u(t)-\partial_t v(t)$, we conclude that \begin{align*} u=v \quad \text{in } C([0,T];H^1_0(U)) \end{align*} and \begin{align*} \partial_t u=\partial_t v \quad \text{in } C([0,T];L^2(U)). \end{align*} This is the desired uniqueness statement. [/step]